# How do you find the local max and min for f(x)=x+(1/x)?

Mar 6, 2017

Local ${f}_{\min} = \left(1 , 2\right)$
Local ${f}_{\max} = \left(- 1 , - 2\right)$

#### Explanation:

$f \left(x\right) = x + \frac{1}{x}$

$f \left(x\right)$ will have extrema where $f ' \left(x\right) = 0$

$f ' \left(x\right) = 1 - \frac{1}{x} ^ 2$ [Power rule]

$1 - \frac{1}{x} ^ 2 = 0 \to {x}^{2} = 1$

$x = \pm 1$

As can be seen by the graph of $f \left(x\right)$ below, $f \left(x\right)$ has a local maximum at $x = - 1$ and a local minimum at $x = + 1$

graph{x+1/x [-12.66, 12.65, -6.33, 6.33]}

Therefore:
Local ${f}_{\min} = f \left(1\right) = 1 + 1 = 2 \to \left(1 , 2\right)$
and
Local ${f}_{\max} = f \left(- 1\right) = - 1 - 1 = - 2 \to \left(- 1 , - 2\right)$